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Elastic Collision of two balls

  1. Jun 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Two balls, Ball A on the left, and Ball B on the right, are moving towards each other. Ball A is much bigger and heavier than Ball B. Ball A is moving with speed x, and Ball B is moving with speed y. They collide, and an elastic collision occurs, sending Ball B rebounding backwards.
    Diagrams:
    Before: A---> <---B
    After: A(unknown direction,probably still to the right) B---->

    Terms:
    Ball B final speed: Bf
    Ball A final speed: Af
    Ball B initial speed: Bi = Y
    Ball A initial speed Ai = X

    Question: What is Ball B's speed after the collision? (Meaning solve for Bf)


    2. Relevant equations
    Assume perfectly elastic collision
    (1/2)MV1i^2+(1/2)MV2i^2 = (1/2)MV^2 + (1/2)MV^2 (conservation of total kinetic energy)

    MV+MV=MV+MV (conservation of total momentum)

    Coefficient of restitution = velocity of separation/velocity of approach
    Coefficient of restitution = Bf-Af/Ai-Bi


    3. The attempt at a solution
    Since no numbers are given, the answer is in terms of symbols. I know that since it's elastic collision, the coefficient of restitution is 1. So,
    [tex]\frac{B_{f}-A_{f}}{A_{i}-B_{i}} = \frac{B_{f}-A_{f}}{X-(-Y)} =1[/tex]
    So,
    [tex]B_{f}-A_{f} = X+Y[/tex]
    I want to solve for Bf, but I don't know Af. That's the problem.
    Maybe Af stays the same as Ai (meaning Ball A doesn't change speed)?
    Using the conservation of momentum equation is difficult because there are no masses given, so that will just complicate it even more.
    So what do I do?
     
    Last edited: Jun 25, 2007
  2. jcsd
  3. Jun 25, 2007 #2

    Doc Al

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    Here's a hint: You are free to use any reference frame you like to view the problem. There may be a frame in which the answer is obvious. (You can always transform back to the original lab frame when done.)
     
  4. Jun 25, 2007 #3

    nrqed

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    there is a problem right here. You are using the same mass for A and B which is not the case, if I understand your question. Also, you are using the same speed for both of them after the collision, which is not the case.

    Again, you are using the same mass for both particles, and you are in addition assuming the same speed for both of them before *and* after. This is not right. You have to rewrite your equations completely.
     
  5. Jun 25, 2007 #4
    Doc Al:
    Thanks for the reply. I don't quite understand what you mean. Should I see the problem from Ball A's point of view, and then use relative speeds? How would I go from there?

    nrqed:
    Thanks for the reply, sorry for not being clear. The equations I listed in the "relevant equations" part were just general forms of the kinetic energy and momentum equations. I forgot to put the right subscripts on them since I didn't actually use them. (I don't know if I have to?) And you're right, the masses are not the same, which is part of the problem.

    Conservation of Momentum:
    [tex]M_{A}A_{i}+M_{B}B_{i}=M_{A}A_{f}+M_{B}B_{f}[/tex]

    Conservation of Kinetic Energy:
    [tex]\frac{1}{2}M_{A}A^2_{i}+\frac{1}{2}M_{B}B^2_{i}=\frac{1}{2}M_{A}A^2_{f}+\frac{1}{2}M_{B}B^2_{f}[/tex]

    Terms:
    Ball B final speed: [tex]B_{f}[/tex]
    Ball A final speed: [tex]A_{f}[/tex]
    Ball B initial speed: [tex]B_{i} = Y[/tex]
    Ball A initial speed [tex]A_{i} = X[/tex]

    Coefficient of restitution = velocity of separation/velocity of approach
    Coefficient of restitution = [tex]\frac{B_{f}-A_{f}}{A_{i}-B_{i}}[/tex]
     
    Last edited: Jun 25, 2007
  6. Jun 25, 2007 #5

    nrqed

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    In principle all you have to do is to isolate A_f from the momentum equation, plug into the kinetic energy equation and then isolate Bf. You then have Bf in terms of the masses and of the initial velocities. This can be simplified if you assume that the mass of A is much larger than the mass of B. It's not clear to me that you can indeed make that approximation.
     
  7. Jun 25, 2007 #6

    Doc Al

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    Yes, I would view the collision from a frame in which Ball A is at rest. The way I interpret the problem, you are to assume that Ball A is MUCH more massive than Ball B. I picture one as a bowling ball, the other as a ping-pong ball.

    Done correctly, this will allow you to do this problem "in your head". (Almost.)

    Doing it the hard way works too! Take heed of nrqed's comments if you pursue that route.
     
  8. Jun 25, 2007 #7
    Doc Al:
    If I view it in a frame with Ball A at rest, then that means Ball B bounces back with speed X-Y (maybe absolute value of X-Y). Is this correct? Do I have to "tranform" it once I come back to the lab reference view?

    nrqed:
    I haven't ignored your post, I'm still working on getting this the reference frame way.
     
  9. Jun 25, 2007 #8

    Doc Al

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    No. What's the relative speed of B w.r.t. A? Note that X and Y are speeds, not velocities.

    Sure.
     
  10. Jun 25, 2007 #9
    The relative speed of B w.r.t. A is Y-X (or is it -Y-X?)
     
  11. Jun 25, 2007 #10

    Doc Al

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    I'd say neither.

    If you and I both run towards each other at 10 mph (with respect to the ground), what is our relative speed?
     
  12. Jun 25, 2007 #11
    Sorry if I seem a bit dense.

    10+10=20 mph

    So would that mean X+Y?
     
    Last edited: Jun 25, 2007
  13. Jun 25, 2007 #12

    Doc Al

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    Yes, much better.
     
  14. Jun 25, 2007 #13
    So, if Ball A were not moving, then Ball B would be coming towards it at speed X+Y. Ball B would bounce back at speed X+Y. So X+Y is the final answer? What, if anything, do I have to do once I come back to the lab view?
     
  15. Jun 25, 2007 #14

    Doc Al

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    Think it through, step by step. In the ball A frame, ball B is approaching at a speed of X+Y (directed to the left). After the collision, what's the velocity of ball B?

    Transform that velocity back to the lab frame and you're done.
     
  16. Jun 25, 2007 #15
    Since Ball A is not moving, it has 0 momentum. So, for momentum to be conserved, the speed of Ball B after the collision would be X+Y. Still, not sure how to "transform" that though.
     
  17. Jun 25, 2007 #16

    Doc Al

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    What allows you to get away with this is the realization that a tiny ball hitting a huge ball at rest just bounces back with the same speed in the opposite direction. (A reasonable approximation, given that A is much bigger than B.) I trust this makes sense to you.

    So the initial velocity of ball B is -(X+Y) and the final velocity is +(X+Y).

    Transforming back to the lab frame is easy. Just do the reverse of what you had to do to get to the Ball A frame. In the lab frame, Ball A had a velocity of +X; what did you have to add/subtract to that velocity (and thus to all velocities) to get to a frame where Ball A's speed was zero?
     
  18. Jun 25, 2007 #17
    I had to subtract X, so now I have to add X.

    (X+Y)+X = 2X+Y

    So, 2X+Y is the final answer.

    Edit: Thank you very much.
     
    Last edited: Jun 25, 2007
  19. Jun 25, 2007 #18

    Doc Al

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    And there you have it. :cool:

    Just for the exercise, you should now solve this problem the long way. (Using the conservation laws.)
     
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